# Derivation of the Rayleigh Distribution Equation

The Rayleigh Distribution makes the following assumptions:

- \(\bar{h} \sim \mathcal{N}(\bar{h},\sigma_h^2), \bar{v} \sim \mathcal{N}(\bar{v},\sigma_v^2)\)
- Horizontal and vertical dispersion are independent.
- \(\sigma_h = \sigma_v\) (realistically \(\sigma_h \approx \sigma_v\))
- \(\rho = 0\)
- No Fliers

for which the PDF is given by:

\(PDF_{r_i}(r) = \frac{r}{\sigma^2}e^{-r^2/2\sigma^2}\)

The PDF can be derived in the following two methods.

# Simplification of the Bivariate Normal Distribution

Using the assumptions above, the Rayleigh Distribution is easily simplified from the Bivariate Normal Distribution which has the equation:

\( f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(h-\bar{h})^2}{\sigma_h^2} + \frac{(v-\bar{v})^2}{\sigma_v^2} - \frac{2\rho(h-\bar{h})(v-\bar{v})}{\sigma_h \sigma_v} \right] \right) \)

By substituting \(\rho = 0\) the equation reduces to:

\( f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v } \exp\left( -\frac{1}{2}\left[ \frac{(h-\bar{h})^2}{\sigma_h^2} + \frac{(v-\bar{v})^2}{\sigma_v^2} \right] \right) \)

Since \(\sigma_h\) and \(\sigma_v\) are equal, substitute \(\sigma\) for each, then collect terms in the exponential, after which the equation reduces to:

\( f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( -\left[ \frac{(h-\bar{h})^2 + (v-\bar{v})^2}{2\sigma^2} \right] \right) \)

Letting \(r^2 = (h-\bar{h})^2 + (v-\bar{v})^2\) the equation becomes:

\( f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)

By transforming to the polar coordinate system one has:

\( f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)